Question

An LCcircuit containing an 80.0mH inductor and a 1.25nF capacitor oscillates with a maximum current of 0.750 A. Assuming the capacitor had its maximum charge at time t= 0, calculate the energy stored in the inductor after 2.50 ms of oscillation.

Short answer

The energy stored in the inductor after 2.50 ms of oscillation is 0.0212J.

Step-by-step solution

Energy conservation in LC circuit

The stored energy in capacitor in addition to the stored energy in inductor at any instant equals to the maximum stored energy in the capacitor.

Mathematically it can be given by

$$\begin{gathered} {\mathrm{U}}_{\mathrm{L}}+{\mathrm{U}}_{\mathrm{C}}={\mathrm{U}}_{\mathrm{C}\max } \\ \frac{1}{2}{\mathrm{Li}}^2+\frac{{\mathrm{q}}^2}{2\mathrm{C}}=\frac{{\mathrm{Q}}^2}{2\mathrm{C}} \end{gathered}$$

Where , ${\mathrm{U}}_{\mathrm{C}\max }$is the maximum energy stored in capacitor and ${\mathrm{U}}_{\mathrm{L}}$and ${\mathrm{U}}_{\mathrm{C}}$are the energy stored in inductor and capacitor respectively.

The definition of oscillating frequency of the LC circuit

It is frequency at which current or charge move in whole LC circuit.

Mathematically it is given

$$\begin{gathered} f=\frac{1}{2\mathrm{\pi }\sqrt{\mathrm{LC}}} \\ \omega =\frac{1}{\sqrt{\mathrm{LC}}} \end{gathered}$$

Where, $L$is the inductance of inductor, $C$is the capacitance of capacitor , $f$is frequency of oscillation and is the angular frequency .

The current at a particular time in LC circuit

The current at particular time is given by

$i=-\omega Q\sin \left(\omega t\right)$

Where, $i$is the current in the circuit at time , $Q$is the maximum charge over capacitor .

Step 2:  Calculation of maximum charge on the capacitor in LC circuit

Using

$\frac{1}{2}{\mathrm{Li}}^2+\frac{{\mathrm{q}}^2}{2\mathrm{C}}=\frac{{\mathrm{Q}}^2}{2\mathrm{C}}$

For maximum current in LC circuit stored charge on capacitor must be zero .

So,

$$\begin{gathered} \frac{1}{2}\mathrm{Li}{\left({\mathrm{i}}_{\max }\right)}^2=\frac{{\mathrm{Q}}^2}{2\mathrm{C}} \\ \mathrm{Q}={\mathrm{i}}_{\max }\sqrt{\mathrm{LC}} \end{gathered}$$

Put the values of constants in above equation

$$\begin{gathered} \mathrm{Q}=\left(0.750\mathrm{A}\right)\sqrt{0.0800\mathrm{H}\times 1.25\times {10}^{-9}\mathrm{F}} \\ \mathrm{Q}=7.50\times {10}^{-6}\mathrm{C} \end{gathered}$$

$7.50\times {10}^{-6}\mathrm{C}$

Thus, the maximum charge on the capacitor is .

Calculation of angular frequency of oscillation

Using

$\omega =\frac{1}{\sqrt{LC}}$

Put the values of constants in above equation

$$\begin{gathered} \mathrm{\omega }=\frac{1}{\sqrt{0.0800\mathrm{H}\times 1.25\times {10}^{-9}\mathrm{F}}} \\ \mathrm{\omega }=1.00\times {10}^5\mathrm{rad}/\mathrm{s} \end{gathered}$$

$1.00\times {10}^5\mathrm{rad}/\mathrm{s}$

Thus, the angular frequency of oscillation .

$t=2.50\mathrm{ms}$

Calculation of current in the LC circuit at a time

Using

$i=-\omega Q\sin \left(\omega t\right)$

Now put the value of constant in above equation

$$\begin{gathered} \mathrm{i}=-\left(1.00\times {10}^5\mathrm{rad}/\mathrm{s}\right)\left(7.50\times {10}^{-6}\mathrm{C}\right)\sin \left(1.00\times {10}^5\mathrm{rad}/\mathrm{s}\times 2.50\times {10}^{-3}\mathrm{s}\right) \\ \mathrm{i}=0.7279\mathrm{A} \end{gathered}$$

Thus, the current in the LC circuit at time 2.50ms is 0.7279A .

Calculation of the energy stored in the inductor after 2.50 ms of oscillation

Using

${\mathrm{U}}_{\mathrm{L}}=\frac{1}{2}\mathrm{Li}$

Now put the values of constants in above equation

$$\begin{gathered} {\mathrm{U}}_{\mathrm{L}}=\frac{1}{2}\left(0.0800\mathrm{H}\right){\left(0.7279\mathrm{A}\right)}^2 \\ {\mathrm{U}}_{\mathrm{L}}=0.0212\mathrm{J} \end{gathered}$$

Hence ,the energy stored in the inductor after 2.50 ms of oscillation is 0.0212J.